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natima [27]
4 years ago
13

Three consecutive even numbers have a sum where one half of that sum is between 90 and 105

Mathematics
1 answer:
Marrrta [24]4 years ago
8 0
2n;\ 2n+2;\ 2n+4-three\ consecutive\ even\ numbers\\\\90 \ \textless \  \dfrac{1}{2}(2n+2n+2+2n+4) \ \textless \  105\\\\90 \ \textless \  \dfrac{1}{2}(6n+6) \ \textless \  105\\\\90 \ \textless \  3n+3 \ \textless \  105\ \ \ |divide\ both\ sides\ by\ 3\\\\30 \ \textless \  n+1 \ \textless \  35\ \ \ |subtract\ 1\ from\ both\ sides\\\\29 \ \textless \  n \ \textless \  34\to n\in\{30;\ 31;\ 32;\ 33\}\\\\Answer:\\60;\ 62;\ 64\ or\ 62;\ 64;\ 66\ or\ 64;\ 66;\ 68\ or\ 66;\ 68;\ 70.
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Given:

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Solution:

The two column proof is:

Statements                                             Reasons

1. Q is between P and R                         1. Given

2. PQ+QR=PR                                  2. Segment addition postulate

3. R is between Q and S                        3. Given

4. QR+RS=QS                                   4. Segment addition postulate

5. PR=QS                                            5. Given

6. PQ+QR=QR+RS                        6. Substituting property of equality

7. PQ+QR-QR=QR+RS-QR    7. Subtraction property of equality

8. PQ=RS                                            8. Simplify

Hence proved.

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